[sura01]

solid Ca(OH)2 of 0.1 M of KOH was mixed.phenolphalein was used. The burette contained 0.13 M H2SO4

through titration i got 11.0 cm3.

can someone help me to find the:
1)number of moles of KOH in 25 ml of 0.1M KOH
2)number of moles of H2SO4 that will react with the moles of KOH
3)From titration results, determine the number of moles of 0.13M H2SO4 used
4)the number of moles of H2SO4 that reacted with Ca(OH)2
5)number of moles of Ca(OH)2 present in 25 ml of solution
6)the concentration(solubility) of Ca(OH)2 in 1 dm3 of solution
7)the solubility product of Ca(OH)2

Any help on these would be greatly appreciated.

[sura01]

i was thinking that to calculate the number of moles of KOH in 25 ml of 0.1 would be:
1000 cm= 0.1 M
25 cm = 2.5 * 10^-3 mol KOH.

and for the number of moles of H2SO4 that would react with KOH is:
2 KOH + H2SO4 = K2SO4 + 2H20
2 moles react with 1 mole.
1 mole/ 2 mol * 2.5 * 10^-3 = 1.25 * 10-3 mole H2SO4.

thats all i really understand, can someone help me out with the rest and tell me if what i did so far is correct? thanks!

[Borek]

OK so far, why don't you use the same method you used in 1 to calculate number of moles of sulfuric acid?

[sura01]

so it be :
1000 cm= 0.13 ml
11.0 cm = 1.43 * 10^-3 moles of H2SO4

is it correct?
and how about finding the number of moles of H2SO4 that would react with Ca(OH)2? i mean how would i go about finding that? please *delete me*

[Borek]

Acid reacted with both KOH and Ca(OH)₂. Write both reaction equations - how much acid reacted with KOH? This was a simple stoichiometry. Now it is a time for a simple subtraction - how much acid was left for Ca(OH)₂?